Given an array arr[] of size N, the task is to find the number of operations to convert array elements to zero by decrementing the value of array elements in pairs by any positive value. If the array elements can’t be converted to 0, return -1.
Input: arr[] = {3, 2}
Output: -1
Explanation: All the array elements can’t be converted to 0
Input: arr[] = {5, 4, 3}
Output: 12
Explanation: Subtract 1 from pair (4, 3) we get {5, 3, 2}, subtract 3 from (5, 3) we get {2, 0, 2}, Subtract 2 from pair (2, 2) we get {0, 0, 0}
Approach: The task can be solved by storing all elements of an array in a priority queue, Then we need to choose pair of the two greatest elements from the queue and subtract 1 from them until only one or no positive element is left.
Follow the below steps to solve the problem:
- Store elements in the priority queue
- Take a while loop until the size of the priority queue is greater than or equal to 2, and in each iteration:-
- Pop the first two elements and store them in variables ele1 and ele2
- Decrement ele1 and ele2 with 1, If any of them are still greater than zero push them again in the queue.
- If the queue is empty print the number of operations needed, else -1
Below is the implementation of the above algorithm:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to check whether // all element of vector can // become zero after the operations void gfg(vector< int >& v)
{ // Priroty queue to store
// elements of vector v
priority_queue< int > q;
// Loop to store elements
// in priroty queue
for ( auto x : v) {
q.push(x);
}
// Stores the number
// of operations needed
int cnt = 0;
while (q.size() >= 2) {
// Variable to store greatest
// element of priority queue
int ele1 = q.top();
q.pop();
// Variable to store second greatest
// element of priority queue
int ele2 = q.top();
q.pop();
// Decrementing both by 1
ele1--;
ele2--;
cnt += 2;
// If elements are greater
// then zero it is again
// stored in the priority queue
if (ele1) {
q.push(ele1);
}
if (ele2) {
q.push(ele2);
}
}
if (q.size() == 0)
cout << cnt << endl;
else
cout << -1;
} // Driver code int main()
{ vector< int > v = { 5, 3, 4 };
gfg(v);
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to check whether
// all element of vector can
// become zero after the operations
static void gfg( int [] v)
{
// Priroty queue to store
// elements of vector v
PriorityQueue<Integer> q = new PriorityQueue<>(Collections.reverseOrder());
// Loop to store elements
// in priroty queue
for ( int x : v) {
q.add(x);
}
// Stores the number
// of operations needed
int cnt = 0 ;
while (q.size() >= 2 ) {
// Variable to store greatest
// element of priority queue
int ele1 = q.peek();
q.remove();
// Variable to store second greatest
// element of priority queue
int ele2 = q.peek();
q.remove();
// Decrementing both by 1
ele1--;
ele2--;
cnt += 2 ;
// If elements are greater
// then zero it is again
// stored in the priority queue
if (ele1> 0 ) {
q.add(ele1);
}
if (ele2> 0 ) {
q.add(ele2);
}
}
if (q.size() == 0 )
System.out.print(cnt + "\n" );
else
System.out.print(- 1 );
}
// Driver code
public static void main(String[] args)
{
int [] v = { 5 , 3 , 4 };
gfg(v);
}
} // This code is contributed by shikhasingrajput |
# Python code for the above approach from queue import PriorityQueue
# Function to check whether # all element of vector can # become zero after the operations def gfg(v):
# Priroty queue to store
# elements of vector v
q = PriorityQueue()
# Loop to store elements
# in priroty queue
for i in range ( len (v)):
q.put( - 1 * v[i])
# Stores the number
# of operations needed
cnt = 0
while (q.qsize() > = 2 ):
# Variable to store greatest
# element of priority queue
ele1 = - 1 * q.get()
# Variable to store second greatest
# element of priority queue
ele2 = - 1 * q.get()
# Decrementing both by 1
ele1 = ele1 - 1
ele2 = ele2 - 1
cnt = cnt + 2
# If elements are greater
# then zero it is again
# stored in the priority queue
if ele1 > 0 :
q.put( - 1 * ele1)
if ele2 > 0 :
q.put( - 1 * ele2)
if q.qsize() = = 0 :
print (cnt)
else :
print ( - 1 )
# Driver code v = [ 5 , 3 , 4 ]
gfg(v) # This code is contributed by Potta Lokesh |
// C# program for the above approach using System;
using System.Collections.Generic;
public class GFG{
// Function to check whether
// all element of vector can
// become zero after the operations
static void gfg( int [] v)
{
// Priroty queue to store
// elements of vector v
List< int > q = new List< int >();
// Loop to store elements
// in priroty queue
foreach ( int x in v) {
q.Add(x);
}
// Stores the number
// of operations needed
int cnt = 0;
while (q.Count >= 2) {
// Variable to store greatest
// element of priority queue
int ele1 = q[0];
q.RemoveAt(0);
// Variable to store second greatest
// element of priority queue
int ele2 = q[0];
q.RemoveAt(0);
// Decrementing both by 1
ele1--;
ele2--;
cnt += 2;
// If elements are greater
// then zero it is again
// stored in the priority queue
if (ele1 > 0) {
q.Add(ele1);
}
if (ele2 > 0) {
q.Add(ele2);
}
}
if (q.Count == 0)
Console.Write(cnt + "\n" );
else
Console.Write(-1);
}
// Driver code
public static void Main(String[] args)
{
int [] v = { 5, 3, 4 };
gfg(v);
}
} // This code is contributed by 29AjayKumar |
<script> // javascript program for the above approach // Function to check whether
// all element of vector can
// become zero after the operations
function gfg(v) {
// Priroty queue to store
// elements of vector v
q = [];
// Loop to store elements
// in priroty queue
for (x of v) {
q.push(x);
}
q.sort((a, b)=>b - a); // Stores the number
// of operations needed
var cnt = 0;
while (q.length >= 2) {
// Variable to store greatest
// element of priority queue
var ele1 = q[0];
q.splice(0, 1);
// Variable to store second greatest
// element of priority queue
var ele2 = q[0];
q.splice(0, 1);
// Decrementing both by 1
ele1 -= 1;
ele2 -=1;
cnt += 2;
// If elements are greater
// then zero it is again
// stored in the priority queue
if (ele1 > 0) {
q.push(ele1);
}
if (ele2 > 0) {
q.push(ele2);
}
}
if (q.length == 0)
document.write(cnt + "\n" );
else
document.write(-1);
}
// Driver code
var v = [ 5, 3, 4 ];
gfg(v);
// This code is contributed by umadevi9616 </script> |
12
Time Complexity: O(N)
Auxiliary Space: O(N)